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**Hint:**Rate of a reaction depends upon the concentration of substrate or product. For the second order reaction,

Rate = k${\left[ A \right]^2}$

where k= rate constant and A= concentration

The units for rate are $M{s^{ - 1}}$

_{.}

**Complete step by step answer:**

Rate of a reaction is the change in concentration of reactant or product with time.

First, we will see what is given to us in the question.

So, we have rate= 0.30$M{s^{ - 1}}$ when the concentration=0.20M

Let this rate be ${r_1}$and concentration be ${c_1}$.

Thus, ${r_1}$=k${\left[ A \right]^2}$

0.30$M{s^{ - 1}}$=k$ \times $${\left[ {0.20} \right]^2}$M

Or k =$\dfrac{{0.30M{s^{ - 1}}}}{{{{\left[ {0.20} \right]}^2}M}}$ Let this be equation 1.

We have to find out the rate when the concentration is three times the initial concentration (${c_1}$).

Let this rate be ${r_2}$ and concentration be ${c_2}$.

We know rate (${r_2}$)= k${\left[ A \right]^2}$

We concentration (${c_2}$)=$3 \times {c_1}$

${c_2}$=3$ \times $0.20M

${c_2}$=0.60M =[A]

We need the value of k so that we can find the rate. So, this we will find from equation 1.

From equation 1, we have

k =$\dfrac{{0.30M{s^{ - 1}}}}{{{{\left[ {0.20} \right]}^2}M}}$

k=$\dfrac{{0.30M{s^{ - 1}}}}{{0.04M}}$

k=7.5${s^{ - 1}}$

Now, we have got the value of k. We can put other values and get our answer as

rate (${r_2}$)= k${\left[ A \right]^2}$

${r_2}$=$7.5{s^{ - 1}} \times {[0.60]^2}M$

${r_2}$=$7.5{s^{ - 1}} \times 0.36M$

${r_2}$=2.7$M{s^{ - 1}}$

So, the correct answer is option (d).

**Note:**

1)Always write units during numericals. It decreases the chances of mistakes.

2)Read the question carefully, from question only we got to know that the reaction is second order and we have used formula accordingly. There is a different formula for first, second and third-order reactions. For example,

For First order reaction- rate=${k_1}{\left[ A \right]^1}$

For Second order reaction- rate= ${k_2}{\left[ A \right]^2}$

For Third order reaction- rate= ${k_3}{\left[ A \right]^3}$

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